Question

A solution is prepared by adding 50.0 mL of 0.040 M HClO4 and 50.0 mL of 0.060 M HCl.(a) Calculate the concentrations of all ions in solution.(b) Calculate the pH of this solution.

Answer 1

Part a)

Here,

Volume of HClO₄ = 50 mL = 0.05 L

Volume of HCl = 50 mL = 0.05 L

Moles of HClO₄ = Volume x Concentration of HClO₄ = 0.05 L x 0.040 M = 0.002 mol

Similarly,

Moles of HCl = Volume x Concentration of HCI = 0.05 L x 0.06 = 0.003 mol

Here, total volume of solution = 50 + 50 = 100 mL = 0.1 L

HClO₄ H⁺ + ClO₄^-

HCI H⁺ + CI^-

Therefore,

Moles of H⁺ = Moles of HClO₄ + Moles of HCI = 0.002 + 0.003 = 0.005 mol

Concentration of H⁺ = Moles of H⁺/Total volume = 0.005/0.1 = 0.05 M

Moles of ClO₄^-- = Moles of HClO₄ = 0.002 mol

Concentration of ClO₄^- = Moles of ClO₄^-/Total volume = 0.002/0.1 = 0.02 M

Moles of CI^- = Moles of HCI = 0.003 mol

Concentration of CI^- = Moles of CI^-/Total volume = 0.003/0.1 = 0.03 M

Part b)

We have,

pH = -log[H⁺]

Here, H⁺ = 0.05 M

Therefore, pH = -log (0.05) = 1.30